Total time to reach max height:
\[
t = \frac{u}{g} = \frac{20}{10} = 2\,s
\]
Displacement in last second of upward motion (from \(t=1\) to \(t=2\)):
Displacement from ground at \(t=1\):
\[
s_1 = ut - \frac{1}{2}gt^2 = 20 \times 1 - \frac{1}{2} \times 10 \times 1^2 = 20 - 5 = 15\,m
\]
Displacement at \(t=2\) (max height):
\[
s_2 = 20 \times 2 - \frac{1}{2} \times 10 \times 4 = 40 - 20 = 20\,m
\]
Displacement during last second:
\[
s_2 - s_1 = 20 - 15 = 5\,m
\]